First slide

Mechanical equilibrium force cases

Question

A mass M is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.

Easy

A

$\sqrt{2} M g$
B

$\sqrt{3} M g$
C

$2 M g$
D

$3 M g$

Solution

As there is a load at P, so tension in AP and PB will be different. Let these be T₂ and T₁, respectively. For vertical equilibrium at P,
$T_{1} \cos 60 ° = M g, i.e. T_{1} = 2 Mg . . . . . . . (i)$
and for horizontal equilibrium at P,
$T_{2} = T_{1} \sin 60^{0} = T_{1} (\sqrt{3} / 2)$
Substituting the value of T₁ from Eq. (i), we get
$T_{2} = (2 M g) \times (\sqrt{3} / 2) = \sqrt{3} M g$

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